The area of the region bounded by the curve \(y^2 = 4x\), \(y\)-axis and the line \(y = 2\), is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{2}{3}\) sq. units
The region lies between the y-axis and the parabola, from y = 0 (the vertex) up to the line y = 2. So it is an area with respect to the y-axis, using horizontal strips of length x:
\( \displaystyle x = \frac{y^2}{4} \)
\( \displaystyle \text{Area} = \int_{0}^{2}\frac{y^2}{4}\,dy = \Big[\frac{y^3}{12}\Big]_{0}^{2} \)
\( \displaystyle = \frac{8}{12} = \frac{2}{3} \) sq. units
Why the other options are incorrect
Option (2) \(\displaystyle\frac{4}{3}\) sq. units — Incorrect. This is the area between the parabola and the x-axis from x = 0 to x = 1 (where y = 2), that is the region on the other side of the curve inside the rectangle 1 × 2; it equals 2 − \(\frac{2}{3}\).
Option (3) \(\displaystyle\frac{3}{2}\) sq. units — Incorrect. The integral gives \(\frac{2}{3}\), not \(\frac{3}{2}\).
Option (4) \(\displaystyle\frac{3}{4}\) sq. units — Incorrect. The integral gives \(\frac{2}{3}\), not \(\frac{3}{4}\).